home work on sequences

7/30/2019 Home Work on Sequences

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LIMITS ON SEQUENCES BY DILEEP Page 1

LIMITS ON SEQUENCES:

EXAMPLE 1: Does the following sequence converge or diverge?

SOLUTION: I will first use long division to simplify this rational expression.

Now I will look at the limit.

Since the limit approaches a finite value, we can conclude that the

sequence converges.


SOLUTION: Again, I will use long division to simplify this radical expression.

Now, I will look at the limit of this sequence.


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Since the limit exists, we can conclude that the sequence

converges.


SOLUTION: First let us simplify this rational expression.

Now, determine the limit of this sequence.

Since the limit goes off to infinity, therefore, the sequence

diverges.



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SOLUTION:

Therefore, this sequence does converge.


SOLUTION: How are we going to deal with the fact that we have a cosine in

this sequence? Well, if you can remember when we discussed

limits in Calculus I, you should remember that there is a theorem

called the Sandwich Theorem that we used for functions similar to

this function. Here is the Sandwich Theorem.

FACT:

We now have to determine if the two new sequences converge or

diverge to the same value.

Since both limits converge to the same value, therefore the limit

of a nalso converges to 0 by the sandwich theorem.


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SOLUTION: We know that -1 sin n 1, which implies 0 sin2

n 1,

therefore

Therefore, by the sandwich theorem, the sequence a n converges.

For some of these problems, especially the first few that were discussed, it would have

been nice to be able to use l ' Hpital's Rule, but we could not because we are talking

about sequences and not functions. The next two facts will help us make a connectionbetween sequences and functions, so we will be able to use l ' Hpital's Rule.

FACT: The Continuous Function Theorem for Sequences

Let {a n} be a sequence of real numbers. If a n L and if f is a function

that is continuous at L and defined at all a n, then f (a n) f (L).



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SOLUTION:

How did I simplify b n? I used long division to simplify b n. Now

let us determine if the sequence bn

converges or diverges.

FACT: Suppose that f (x) is a function defined for all x n 0 and

that {a n} is a sequence of real numbers such that a n = f

(n) for n n 0.

This fact is stating that the corresponding function f (x) will have the same limit as its

corresponding sequence, a n.


SOLUTION:

So let us look at the limit of f (x).

(The first limit evaluates to the indeterminate form /.)


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Since the limit exists, we can conclude that a n converges

to zero.

Isn't it nice when we can use l ' Hpital's Rule!


SOLUTION:

Since the limit exists, we can conclude that the sequence

a nconverges to zero.


SOLUTION:

, - is an indeterminate form. I am going to convert this

indeterminate form into the indeterminate form of / by

rationalizing the numerator.


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This is now in the form of /. Now, I am going to divide both

top and bottom by x, taking note to divide everything inside the

radical by x2.

Therefore the sequence a n converges.

For equations 3 - 6, x is number that remains fixed.


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SOLUTION:


SOLUTION:


SOLUTION:

EXAMPLE 14: Does this sequence converge or diverge?

SOLUTION:


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SOLUTION:

So this is in the form of equation 2.

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